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Electric field at the center of an equilateral triangle

  • Thread starter rocapp
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Homework Statement



See attached image

Homework Equations



1/4πε*q/r

The Attempt at a Solution



I know that since the problem gives centimeters, there are initial changes to units to be made in the permittivity constant.

This makes it 8.86x10^-9 N/nanoCoulombs and 100 mm.

But I'm not sure how to apply my electric field equation.

Since it is an equilateral triangle, the two positive charges on either side are the same, so that value is doubled. The angles are 30°. I'm not sure what to do beyond this. Is this a case of the y-values canceling, and then integrating the x-values?

Please help. Thanks in advance!
 

Answers and Replies

  • #2
SammyS
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Homework Statement



See attached image

Homework Equations



1/4πε*q/r

The Attempt at a Solution



I know that since the problem gives centimeters, there are initial changes to units to be made in the permittivity constant.

This makes it 8.86x10^-9 N/nanoCoulombs and 100 mm.

But I'm not sure how to apply my electric field equation.

Since it is an equilateral triangle, the two positive charges on either side are the same, so that value is doubled. The angles are 30°. I'm not sure what to do beyond this. Is this a case of the y-values canceling, and then integrating the x-values?

Please help. Thanks in advance!
It generally makes more sense to change units from cm to meters than to change units of the physical constants.
 
  • #3
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Ok, cool.
So I'm using:

(3.8x10^-8) C = Q

ε = 8.85x10^-12 F/m

The distance from each of the rods can be found using a right triangle and Pythagorean theorem:

√(0.05 m)^2 + (0.1 m)^2

= 0.112 m/2
r = 0.056 m

I'm almost positive that in using the formula for electric field, there must be trigonometry involved to find the net field at the center point. But I'm not sure what to do exactly.

I know that sin(30) = 0.5' = -56.61°

and cos(30) = 0.866' = 8.84°

I know also that the charges must cancel to some extent.
 
Last edited:
  • #4
SammyS
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I forgot to mention, that I don't see any image, so I really don't know any details of the problem you're trying to solve.
 
  • #5
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Ah. There are three uniformly charged rods arranged in an equilateral triangle. Each is 10 cm in length. Two on either side have a charge of 38nC. The bottom one has a charge of -38nC. What is the net electric field on the point Q in the center of the triangle?
 
  • #6
SammyS
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Ah. There are three uniformly charged rods arranged in an equilateral triangle. Each is 10 cm in length. Two on either side have a charge of 38nC. The bottom one has a charge of -38nC. What is the net electric field on the point Q in the center of the triangle?
Do you know how to find the electric field along the perpendicular bisector of a finite length uniform line charge ? -- In this case that's along the perpendicular bisector of a 10 cm rod with uniform charge, 38nC .
 

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